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All components of a quantum system are always entangled. It's not some mysterious property that arises under unusual circumstances. In fact, I would argue tha

All components of a quantum system are always entangled. It's not some mysterious property that arises under unusual circumstances. In fact, I would argue that quantum entanglement is the defining feature of quantum mechanics that distinguishes it from classical mechanics. However, the way that most articles (and even physicists) use the term "entanglement" is in a loose way. Within a good approximation, a multi-component quantum state can be represented as a tensor product of single component quantum states. When this isn't a good approximation anymore, you can say that the entanglement of the system is much more apparent. To give an example of what I mean, there's a technique in quantum chemistry known as density functional theory (DFT), which is used to compute ground state energies of various molecules. For some molecules, it works pretty well. The benefit of DFT is that it is a fast calculation technique (well, as far as quantum chemistry goes), but it's speed comes at a price. Rather than using the Coulombic interaction of every pair of electrons to compute the energy of a molecule (or more technically, using every Coulombic interaction as terms in the Hamiltonian), a probability cloud of electron positions is computed instead, and this cloud is used as the energetic term. This works really well for a lot of systems, but in some cases (superconductors, metals, solid-state physics, van der Waals interactions), this approximation falls apart because the effect of electron-electron correlation significantly affects the energy of the system. In this case, the full, non-separable wavefunction is required. The fact that this full wavefunction cannot easily be broken into an approximation of simpler, single electron wavefunctions means that the entanglement of the system is very apparent.

orbifold 12y ago
For someone who is more of a mathematician than physicist, the following article might be of interest in this context: http://math.ucr.edu/home/baez/quantum/node4.html http://math.ucr.edu/home/baez/quantum/node4.html The gist is that the state space of a multicomponent quantum mechanical system (think electrons in a metal) is given by the tensor product of the single component state spaces. As might be familiar from linear algebra, not every element in the tensor product of two vector spaces is simply a tensor product of two vectors. That's really all there is to entanglement, but unfortunately you can't explain that to a lay person.
gus_massa 12y ago
I agree completely. Entanglement is everywhere. To calculate the correct energy of the electrons in a molecule it's necessary to use entanglement. But these molecular examples may be explained using some weird classical models with local hidden variables, and with electrons that interact with each other to conspire and provide the right result of the measurement. The quantum mechanical explanation is actually more simple, but not "intuitive". Almost all the discussions about entanglement discuss the strange case of two entangled particles that are far away. The distance between the particles is only a trick to: * be sure that one of them can't communicate to the other and tell the result of the measurement * be sure that there is no a local hidden variable theory that explain the result
- musername 12y ago
  why one hidden variable, when you could have one for each atom between the entangled? :)
bmnmasdas 12y ago
I would argue that using the term "entanglement" to refer to the situation where "the entanglement of the system is much more apparent" or "tensor products are not a good approximation anymore" is completely reasonable. In contrast, when you say that "all components of a quantum system [that appears in nature] are always entangled" then this is trivially true in a technical sense, but highly misleading. To produce and stabilize non-trivial amounts of entanglement in a way that it can be harnessed for quantum information processing is certainly an interesting and highly non-trivial task which people are spending lots of effort on.
- Xcelerate 12y ago
  I don't think it's trivially true in a technical sense at all. It certainly didn't have to be the case that all quantum systems exhibit entanglement. Physics could have been different, such that entanglement is a property only some systems can have. And that would lead to drastically different conclusions. In fact, treating entanglement as something special is, at least for me personally, the highly misleading concept. When I was learning quantum mechanics, there were many concepts that confused and bothered me because people explained things in a loose and inaccurate way. Other incorrect statements that confused me for a long time include: "Fermions cannot occupy the same quantum state. Bosons can." "It's not possible to measure the position and momentum of a particle simultaneously." And while not QM: "Mass can be converted to energy." Essentially, schools teach the approximations first, which leads you to believe that reality works a certain way, and then you think of all sorts of situations where these rules lead to paradoxes and contradictions, and then you have to relearn everything the more accurate way, while trying to expunge the old, wrong stuff from your brain that likes to stick there. In my undergrad, I thought of plenty of ways that "entanglement", based upon the way it was explained in class could be used to transfer information faster than the speed of light. The professor didn't have a rebuke against my argument (granted this was a chemistry professor, not a physics professor), and essentially it was because everyone was repeating wrong, catchy phrases to each other. I don't think "Entanglement is the fact that a multiparticle wavefunction cannot be decomposed" is a particularly complex idea to understand. Everything else kind of falls out of that statement. In my opinion, the most accurate theory should be taught first, and then the less accurate approximations can be derived as limiting cases of the more accurate formulation. And if the math is too complex for the more accurate version, then clearly specify what fallacies or assumptions are being taken. One of my statistical mechanics books does this really, really well (the one by McQuarrie). Disclaimers are all throughout his book about how the equations apply to limiting cases and approximations, what those approximations are, and how using those approximations causes the result to differ from a more accurate theory.
lisper 12y ago
> All components of a quantum system are always entangled. No, that's not true. If it were then there could never be direct observation of quantum interference because entangled particles do not self-interfere. It is true that entanglement is very common, and an unentangled state is the unusual case. But it's not true that unentangled states (or, to be precise, unentangled degrees of freedom) don't exist.
- Xcelerate 12y ago
  I think either you're confused on an issue or I'm confused on what you're trying to say. Quantum interference is observable with even a single particle. > because entangled particles do not self-interfere Entangled particles certainly can, and do interfere (as mentioned, electrons in an atom are entangled and definitely interfere; there's a whole field devoted to calculating this effect), so I'm not sure exactly what you mean here. Can you give me, in ket notation, an example of an unentangled state? (And if you're referring to a direct tensor product of two single particle states, I'm still considering that an entangled state since it can unitarily evolve out of that configuration into one that can't be written as a tensor product).
  - bmnmasdas 12y ago
    > (And if you're referring to a direct tensor product of two single particle states, I'm still considering that an entangled state since it can unitarily evolve out of that configuration into one that can't be written as a tensor product) This reasoning does not make sense. The notion of entanglement is not invariant under "global" operations, nor should it be.
  - lisper 12y ago
    > Quantum interference is observable with even a single particle. Yes, but only if that particle is in an unentangled state. If that particle is entangled then the only way to observer the interference is to make measurements on both the particle whose interference you wish to observe and the particle with which it is entangled. This is simply the old canard about measurement "destroying" interference (because measurement and entanglement are the same thing). > Entangled particles certainly can, and do interfere That's true, but the way in which they interfere is different, and the procedure you need to perform to observe the interference is likewise different. > Can you give me, in ket notation, an example of an unentangled state? |U> + |D> Or if you want two unentangled particles: (|U1> + |D1>)(|U2> + |D2>) as opposed to the entangled state: |U1>|D2> + |D1>|U2> (Where U and D are intended to designate something like spin Up or spin Down, but that doesn't really matter.) > if you're referring to a direct tensor product of two single particle states, I'm still considering that an entangled state since it can unitarily evolve out of that configuration into one that can't be written as a tensor product That's news to me. How does that happen?
chm 12y ago
Hmmm... DFT is a method. It has been used by physicists for the last 40 years to model the solid state. What you put forward in your third paragraph is not quite right - things are more subtle. What you seem to be suggesting is that we should be free of the Kohn-Sham framework[1,2] or find the exact XC functional. Not surprisingly, the results of calculations depend on how the functionals were constructed. But that doesn't mean "DFT" is bad. DFT's just the method. There is no approximation in DFT. It is in itself exact for the description of the ground state. [1]:https://www.princeton.edu/mae/people/faculty/carter/ecdocs/EAC-107.pdf https://www.princeton.edu/mae/people/faculty/carter/ecdocs/E... [2]: http://scitation.aip.org/content/aip/journal/jcp/127/2/10.1063/1.2747249 http://scitation.aip.org/content/aip/journal/jcp/127/2/10.10...
- Xcelerate 12y ago
  Yeah, I know. My rant further down the page on inaccuracies in teaching is coming back to haunt me now. Ironic, isn't it? I was glossing over specifics and loosely conflating concepts to form a point.

orbifold 12y ago

For someone who is more of a mathematician than physicist, the following article might be of interest in this context: http://math.ucr.edu/home/baez/quantum/node4.html http://math.ucr.edu/home/baez/quantum/node4.html The gist is that the state space of a multicomponent quantum mechanical system (think electrons in a metal) is given by the tensor product of the single component state spaces. As might be familiar from linear algebra, not every element in the tensor product of two vector spaces is simply a tensor product of two vectors. That's really all there is to entanglement, but unfortunately you can't explain that to a lay person.

gus_massa 12y ago

I agree completely. Entanglement is everywhere. To calculate the correct energy of the electrons in a molecule it's necessary to use entanglement. But these molecular examples may be explained using some weird classical models with local hidden variables, and with electrons that interact with each other to conspire and provide the right result of the measurement. The quantum mechanical explanation is actually more simple, but not "intuitive". Almost all the discussions about entanglement discuss the strange case of two entangled particles that are far away. The distance between the particles is only a trick to: * be sure that one of them can't communicate to the other and tell the result of the measurement * be sure that there is no a local hidden variable theory that explain the result

musername 12y ago
why one hidden variable, when you could have one for each atom between the entangled? :)

bmnmasdas 12y ago

I would argue that using the term "entanglement" to refer to the situation where "the entanglement of the system is much more apparent" or "tensor products are not a good approximation anymore" is completely reasonable. In contrast, when you say that "all components of a quantum system [that appears in nature] are always entangled" then this is trivially true in a technical sense, but highly misleading. To produce and stabilize non-trivial amounts of entanglement in a way that it can be harnessed for quantum information processing is certainly an interesting and highly non-trivial task which people are spending lots of effort on.

Xcelerate 12y ago
I don't think it's trivially true in a technical sense at all. It certainly didn't have to be the case that all quantum systems exhibit entanglement. Physics could have been different, such that entanglement is a property only some systems can have. And that would lead to drastically different conclusions. In fact, treating entanglement as something special is, at least for me personally, the highly misleading concept. When I was learning quantum mechanics, there were many concepts that confused and bothered me because people explained things in a loose and inaccurate way. Other incorrect statements that confused me for a long time include: "Fermions cannot occupy the same quantum state. Bosons can." "It's not possible to measure the position and momentum of a particle simultaneously." And while not QM: "Mass can be converted to energy." Essentially, schools teach the approximations first, which leads you to believe that reality works a certain way, and then you think of all sorts of situations where these rules lead to paradoxes and contradictions, and then you have to relearn everything the more accurate way, while trying to expunge the old, wrong stuff from your brain that likes to stick there. In my undergrad, I thought of plenty of ways that "entanglement", based upon the way it was explained in class could be used to transfer information faster than the speed of light. The professor didn't have a rebuke against my argument (granted this was a chemistry professor, not a physics professor), and essentially it was because everyone was repeating wrong, catchy phrases to each other. I don't think "Entanglement is the fact that a multiparticle wavefunction cannot be decomposed" is a particularly complex idea to understand. Everything else kind of falls out of that statement. In my opinion, the most accurate theory should be taught first, and then the less accurate approximations can be derived as limiting cases of the more accurate formulation. And if the math is too complex for the more accurate version, then clearly specify what fallacies or assumptions are being taken. One of my statistical mechanics books does this really, really well (the one by McQuarrie). Disclaimers are all throughout his book about how the equations apply to limiting cases and approximations, what those approximations are, and how using those approximations causes the result to differ from a more accurate theory.

lisper 12y ago

> All components of a quantum system are always entangled. No, that's not true. If it were then there could never be direct observation of quantum interference because entangled particles do not self-interfere. It is true that entanglement is very common, and an unentangled state is the unusual case. But it's not true that unentangled states (or, to be precise, unentangled degrees of freedom) don't exist.

Xcelerate 12y ago
I think either you're confused on an issue or I'm confused on what you're trying to say. Quantum interference is observable with even a single particle. > because entangled particles do not self-interfere Entangled particles certainly can, and do interfere (as mentioned, electrons in an atom are entangled and definitely interfere; there's a whole field devoted to calculating this effect), so I'm not sure exactly what you mean here. Can you give me, in ket notation, an example of an unentangled state? (And if you're referring to a direct tensor product of two single particle states, I'm still considering that an entangled state since it can unitarily evolve out of that configuration into one that can't be written as a tensor product).
- bmnmasdas 12y ago
  > (And if you're referring to a direct tensor product of two single particle states, I'm still considering that an entangled state since it can unitarily evolve out of that configuration into one that can't be written as a tensor product) This reasoning does not make sense. The notion of entanglement is not invariant under "global" operations, nor should it be.
- lisper 12y ago
  > Quantum interference is observable with even a single particle. Yes, but only if that particle is in an unentangled state. If that particle is entangled then the only way to observer the interference is to make measurements on both the particle whose interference you wish to observe and the particle with which it is entangled. This is simply the old canard about measurement "destroying" interference (because measurement and entanglement are the same thing). > Entangled particles certainly can, and do interfere That's true, but the way in which they interfere is different, and the procedure you need to perform to observe the interference is likewise different. > Can you give me, in ket notation, an example of an unentangled state? |U> + |D> Or if you want two unentangled particles: (|U1> + |D1>)(|U2> + |D2>) as opposed to the entangled state: |U1>|D2> + |D1>|U2> (Where U and D are intended to designate something like spin Up or spin Down, but that doesn't really matter.) > if you're referring to a direct tensor product of two single particle states, I'm still considering that an entangled state since it can unitarily evolve out of that configuration into one that can't be written as a tensor product That's news to me. How does that happen?

chm 12y ago

Hmmm... DFT is a method. It has been used by physicists for the last 40 years to model the solid state. What you put forward in your third paragraph is not quite right - things are more subtle. What you seem to be suggesting is that we should be free of the Kohn-Sham framework[1,2] or find the exact XC functional. Not surprisingly, the results of calculations depend on how the functionals were constructed. But that doesn't mean "DFT" is bad. DFT's just the method. There is no approximation in DFT. It is in itself exact for the description of the ground state. [1]:https://www.princeton.edu/mae/people/faculty/carter/ecdocs/EAC-107.pdf https://www.princeton.edu/mae/people/faculty/carter/ecdocs/E... [2]: http://scitation.aip.org/content/aip/journal/jcp/127/2/10.1063/1.2747249 http://scitation.aip.org/content/aip/journal/jcp/127/2/10.10...

Xcelerate 12y ago
Yeah, I know. My rant further down the page on inaccuracies in teaching is coming back to haunt me now. Ironic, isn't it? I was glossing over specifics and loosely conflating concepts to form a point.